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Question

Two identical metal plates are given positive charge $${Q}_{1}$$ and $${Q}_{2}$$ $$\left( <{ Q }_{ 1 } \right)$$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $$C$$, the potential difference between them is


A
(Q1+Q2)/2C
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B
(Q1+Q2)/C
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C
(Q1Q2)/C
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D
(Q1Q2)/(2C)
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Solution

The correct option is D $$\left( { Q }_{ 1 }-{ Q }_{ 2 } \right) /(2C)$$

The potential difference between the two identical metal plates is given as

$$C = \dfrac{{{\varepsilon _{\rm{0}}}A}}{d}$$

Let the surface charge density is given as

$${\sigma _1} = {\sigma _{21}} = \dfrac{Q}{A}$$

The net electric field is

$${E_{net}} = \dfrac{{{\sigma _1} - {\sigma _2}}}{{2{\varepsilon _{\rm{0}}}}}$$

We know the potential difference is given as

$$V = E.d$$

By substituting the above values we get

$$V = \dfrac{{{Q_1} - {Q_2}}}{{2C}}$$

 


Physics

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